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16 February, 07:14

If sin x = 1/3 and sec y = 5/4 where x and y lie between 0 and pi/2, evaluate sin (x-y)

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  1. 16 February, 08:18
    0
    sin (x) = ¹/₃

    sin⁻¹[sin (x) ] = sin⁻¹ (¹/₃)

    x ≈ 0.34°

    x = ¹⁷/₉₀₀₀π

    sec (y) = 1¹/₄

    1 = 1¹/₄cos (y)

    ⁴/₅ = cos (y)

    cos⁻¹ (⁴/₅) = cos⁻¹[cos (y) ]

    0.64 ≈ y

    ⁴/₁₁₂₅π = y

    sin (x - y) = sin (¹⁷/₉₀₀₀π - ⁴/₁₁₂₅π)

    sin (x - y) = sin (⁻¹/₆₀₀π)

    sin (x - y) = - sin (¹/₆₀₀π)

    sin (x - y) = - sin (³/₁₀)

    sin (x - y) ≈ - 0.3
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